1. **State the problem:** Eminem will receive 20 payments of 3 million each. The first payment is in 1 year, and the next 19 payments occur every 2 years after that (years 3, 5, 7, ..., 39). The interest rate is 3% annually for the first 4 years, then 10% annually thereafter. We want to find the present value (worth today) of this contract. 2. **Identify the cash flows and discount rates:** - Payments: $3$ million each - Payment times: $t = 1, 3, 5, ..., 39$ - Interest rate $r = 3\%$ for $t \leq 4$ years, then $r = 10\%$ for $t > 4$ 3. **Formula for present value of each payment:** $$PV = \frac{C}{(1+r)^t}$$ where $C$ is the payment, $r$ is the annual interest rate applicable at time $t$, and $t$ is the time in years. 4. **Calculate present value for each payment:** - For $t=1$ (within 4 years), use $r=0.03$ - For $t=3$ (within 4 years), use $r=0.03$ - For $t \geq 5$ (after 4 years), use $r=0.10$ 5. **Calculate present value of payments at $t=1$ and $t=3$:** $$PV_1 = \frac{3}{(1+0.03)^1} = \frac{3}{1.03} \approx 2.9126$$ $$PV_3 = \frac{3}{(1+0.03)^3} = \frac{3}{1.092727} \approx 2.7441$$ 6. **Calculate present value of payments from $t=5$ to $t=39$ every 2 years:** For $n=3$ to $20$, payment at $t=2n-1$ years, discount rate $r=0.10$: $$PV_n = \frac{3}{(1+0.10)^{2n-1}}$$ 7. **Sum the present values:** Total present value: $$PV = PV_1 + PV_3 + \sum_{n=3}^{20} \frac{3}{(1.10)^{2n-1}}$$ 8. **Calculate the sum:** Calculate the sum of the geometric series: Let $a = \frac{3}{(1.10)^5}$ and common ratio $q = \frac{1}{(1.10)^2} = \frac{1}{1.21} \approx 0.8264$ Number of terms $= 18$ (from $n=3$ to $20$) Sum: $$S = a \frac{1 - q^{18}}{1 - q} = \frac{3}{(1.10)^5} \times \frac{1 - (0.8264)^{18}}{1 - 0.8264}$$ Calculate: $$(1.10)^5 = 1.61051$$ $$a = \frac{3}{1.61051} \approx 1.863$$ $$(0.8264)^{18} \approx 0.0423$$ $$S = 1.863 \times \frac{1 - 0.0423}{1 - 0.8264} = 1.863 \times \frac{0.9577}{0.1736} \approx 1.863 \times 5.516 = 10.28$$ 9. **Add all present values:** $$PV = 2.9126 + 2.7441 + 10.28 = 15.9367$$ 10. **Final answer:** The contract is worth approximately $15.94$ million today.